AP Statistics Probability, Random Variables, and Probability Distributions — Worked Answer Explanations

Unit 4 · 12 questions explained

Below is a complete answer key for our AP Statistics Probability, Random Variables, and Probability Distributions practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Probability, Random Variables, and Probability Distributions practice test and come back here to review, or head back to the Probability, Random Variables, and Probability Distributions unit overview.

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Question 1 · Easy
A bag contains 4 red marbles and 6 blue marbles. One marble is drawn at random. What is the probability of drawing a red marble?
- A
  0.40Correct
- B
  0.60
  Why not B: This is the probability of drawing blue, not red.
- C
  0.44
  Why not C: Dividing red by blue (4/6 ≈ 0.67) or some other incorrect operation.
- D
  0.67
  Why not D: Dividing red by blue (4/6 ≈ 0.67) instead of red by total (4/10).
Explanation
$P (red) = \frac{number of red}{total} = \frac{4}{10} = 0.40$ . The total number of marbles is 4 + 6 = 10.

Key takeaway
Classical probability: $P(A) = \frac{\text{outcomes in A}}{\text{total outcomes}}$ when all outcomes are equally likely.
Question 2 · Easy
Events $A$ and $B$ are mutually exclusive with $P (A) = 0.30$ and $P (B) = 0.45$ . What is $P (A or B)$ ?
- A
  0.135
  Why not A: 0.135 = P(A) × P(B); this is the probability of A AND B if they were independent, not the addition rule.
- B
  0.75Correct
- C
  0.775
  Why not C: Using the general addition rule P(A) + P(B) − P(A∩B), but mutually exclusive means P(A∩B) = 0.
- D
  1.00
  Why not D: Adding without the joint probability gives 0.75, not 1.00.
Explanation
Mutually exclusive events cannot both occur, so $P (A \cap B) = 0$ . The addition rule simplifies to $P (A \cup B) = P (A) + P (B) = 0.30 + 0.45 = 0.75$ .

Key takeaway
For mutually exclusive events: $P(A \text{ or } B) = P(A) + P(B)$; there is no overlap to subtract.
Question 3 · Easy
A discrete random variable $X$ has the probability distribution shown. What is $E (X)$ ?

$x$ 1 2 3 4
$P (X = x)$ 0.10 0.30 0.40 0.20
- A
  2.50
  Why not A: 2.5 is the arithmetic average of 1, 2, 3, 4 — ignoring the probabilities.
- B
  2.70Correct
- C
  3.00
  Why not C: 3 is the mode of this distribution, not the expected value.
- D
  1.00
  Why not D: This is the sum of the probabilities, not the expected value.
Explanation
$E (X) = \sum x \cdot P (X = x) = 1 (0.10) + 2 (0.30) + 3 (0.40) + 4 (0.20) = 0.10 + 0.60 + 1.20 + 0.80 = 2.70$ .

Key takeaway
$E(X) = \sum x \cdot P(X = x)$; weight each value by its probability, then sum.
Question 4 · Easy
A student guesses randomly on a 5-question true/false quiz. Let $X$ = number of correct answers. What are the mean and standard deviation of $X$ ?
- A
  $μ = 2.5$ , $σ = 1.25$
  Why not A: 1.25 is the variance $n p (1 - p)$ , not the standard deviation.
- B
  $μ = 2.5$ , $σ \approx 1.12$ Correct
- C
  $μ = 5$ , $σ = 2.5$
  Why not C: Using $n$ for both mean and variance; the mean is $n p = 5 (0.5) = 2.5$ .
- D
  $μ = 0.5$ , $σ = 0.5$
  Why not D: These are the mean and SD of a single Bernoulli trial, not the binomial sum.
Explanation
This is a binomial distribution with $n = 5$ , $p = 0.5$ . $μ = n p = 5 (0.5) = 2.5$ . $σ = n p (1 - p) = 5 (0.5) (0.5) = 1.25 \approx 1.12$ .

Key takeaway
Binomial: $\mu = np$ and $\sigma = \sqrt{np(1-p)}$.
Question 5 · Easy
In a standard deck of 52 cards, two cards are drawn without replacement. Given that the first card drawn is an ace, what is the probability that the second card is also an ace?
- A
  $\frac{4}{52}$
  Why not A: This is the unconditional probability of drawing an ace on the first draw, ignoring the given condition.
- B
  $\frac{1}{17}$ Correct
- C
  $\frac{4}{51}$
  Why not C: Reduces denominator to 51 but keeps 4 aces; after drawing one ace, only 3 remain.
- D
  $\frac{3}{52}$
  Why not D: Reduces the number of aces but forgets to reduce the deck size.
Explanation
After drawing one ace, 3 aces remain in a deck of 51 cards. $P (2nd ace ∣ 1st ace) = \frac{3}{51} = \frac{1}{17} \approx 0.059$ .

Key takeaway
Conditional probability without replacement: update both the numerator (successes remaining) and denominator (total remaining).
Question 6 · Easy
Two independent random variables $X$ and $Y$ have means $μ_{X} = 10$ , $μ_{Y} = 6$ , and standard deviations $σ_{X} = 3$ , $σ_{Y} = 4$ . What are the mean and standard deviation of $X + Y$ ?
- A
  $μ = 16$ , $σ = 7$
  Why not A: Adding standard deviations is incorrect; variances (not SDs) add for independent variables.
- B
  $μ = 16$ , $σ = 5$ Correct
- C
  $μ = 4$ , $σ = 5$
  Why not C: Subtracting the means instead of adding.
- D
  $μ = 16$ , $σ = 25$
  Why not D: Reporting the variance ( $σ^{2} = 9 + 16 = 25$ ) instead of the standard deviation.
Explanation
For independent variables: $μ_{X + Y} = μ_{X} + μ_{Y} = 10 + 6 = 16$ . Variances add: $σ_{X + Y}^{2} = σ_{X}^{2} + σ_{Y}^{2} = 9 + 16 = 25$ , so $σ_{X + Y} = 25 = 5$ .

Key takeaway
Means always add/subtract; variances add (never SDs) for independent variables.
Question 7 · Easy
A geometric random variable $X$ counts the number of trials until the first success, where $p = 0.25$ . What is $P (X = 3)$ ?
- A
  $0.2 5^{3} \approx 0.016$
  Why not A: This omits the two preceding failures; geometric requires (1−p)^(k−1) failures before the success.
- B
  $(0.75)^{2} (0.25) \approx 0.141$ Correct
- C
  $(0.75)^{3} \approx 0.422$
  Why not C: Forgetting to multiply by $p = 0.25$ ; this is the probability of 3 consecutive failures.
- D
  $(1 3) (0.25)^{1} (0.75)^{2} \approx 0.422$
  Why not D: This is the binomial formula for exactly 1 success in 3 trials, not the geometric formula.
Explanation
$P (X = k) = (1 - p)^{k - 1} p$ . For $k = 3$ : $P (X = 3) = (0.75)^{2} (0.25) = 0.5625 \times 0.25 \approx 0.141$ . The first two trials are failures, the third is the first success.

Key takeaway
Geometric PMF: $P(X = k) = (1-p)^{k-1}p$; the first $k-1$ trials are failures, the $k$-th is the first success.
Question 8 · Easy
A candy machine has a 30% chance of dispensing a rare flavor. You try the machine 10 times. Using the binomial model, what is the probability of getting the rare flavor at least once?
- A
  $1 - (0.30)^{10}$
  Why not A: $(0.30)^{10}$ is the probability of always getting the rare flavor; subtracting it does not give the complement of zero.
- B
  $10 \times 0.30 = 0.30$
  Why not B: Multiplying the probability by the number of trials gives the expected value, not the probability of at least one.
- C
  $1 - (0.70)^{10} \approx 0.972$ Correct
- D
  $(0.30)^{1} (0.70)^{9} \approx 0.121$
  Why not D: This is the probability of exactly 1 success, not at least 1.
Explanation
$P (at least one) = 1 - P (none) = 1 - (0.70)^{10} \approx 1 - 0.028 = 0.972$ . The complement rule is the most efficient approach.

Key takeaway
$P(\text{at least one success}) = 1 - P(\text{no successes}) = 1 - (1-p)^n$.
Question 9 · Medium
A student claims that because $P (A) = 0.5$ and $P (B) = 0.4$ , events $A$ and $B$ cannot both occur since $P (A) + P (B) = 0.9 < 1$ . Is this reasoning valid?
- A
  Yes; if $P (A) + P (B) < 1$ , they must be mutually exclusive.
  Why not A: Mutually exclusive events satisfy $P (A \cup B) = P (A) + P (B) \leq 1$ ; this is necessary but not sufficient to conclude they cannot both occur.
- B
  No; $P (A) + P (B) < 1$ does not prevent both events from occurring simultaneously.Correct
- C
  Yes; the sum of individual probabilities must equal 1 for events to overlap.
  Why not C: There is no such requirement; overlapping probabilities are constrained differently.
- D
  No; events can co-occur only if $P (A) + P (B) > 1$ .
  Why not D: Events can co-occur even when $P (A) + P (B) \leq 1$ ; e.g., drawing a red ace has $P (red) + P (ace) \approx 0.5 + 0.077 < 1$ but both can occur.
Explanation
Two events can occur together whenever $P (A \cap B) > 0$ . For example, if $P (A \cap B) = 0.2$ , then $P (A \cup B) = 0.5 + 0.4 - 0.2 = 0.7$ , which is perfectly valid. The sum of individual probabilities being less than 1 is not evidence of mutual exclusivity.

Key takeaway
Events can overlap even when $P(A) + P(B) \leq 1$. Mutual exclusivity means $P(A \cap B) = 0$, not that their probabilities sum to more than 1.
Question 10 · Medium
Events $A$ and $B$ satisfy $P (A) = 0.4$ , $P (B) = 0.5$ , and $P (A \cap B) = 0.2$ . Are $A$ and $B$ independent?
- A
  Yes, because $P (A) + P (B) = 0.9$ and $P (A \cap B) > 0$ .
  Why not A: Independence is not determined by whether $P (A \cap B) > 0$ or by the sum of probabilities.
- B
  Yes, because $P (A \cap B) = P (A) \cdot P (B)$ .Correct
- C
  No, because $P (A \cap B) \neq = 0$ .
  Why not C: Independent events can have a non-zero joint probability; they are not mutually exclusive.
- D
  No, because $P (A ∣ B) \neq = P (A)$ .
  Why not D: $P (A ∣ B) = P (A \cap B) / P (B) = 0.2/0.5 = 0.4 = P (A)$ , so they are indeed independent.
Explanation
Check independence: $P (A) \cdot P (B) = 0.4 \times 0.5 = 0.20 = P (A \cap B)$ . Since the product of marginal probabilities equals the joint probability, $A$ and $B$ are independent. Equivalently, $P (A ∣ B) = 0.2/0.5 = 0.4 = P (A)$ .

Key takeaway
Two events are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$, equivalently $P(A|B) = P(A)$.
Question 11 · Hard
A fair coin is flipped 8 times. Using the binomial distribution, what is the probability of getting exactly 5 heads?
- A
  $\frac{5}{8} = 0.625$
  Why not A: Dividing successes by trials gives a proportion, not a probability from the binomial formula.
- B
  $(5 8) (0.5)^{5} (0.5)^{3} = \frac{56}{256} \approx 0.219$ Correct
- C
  $(0.5)^{8} \approx 0.004$
  Why not C: This is the probability of one specific sequence of 8 outcomes, not all sequences with exactly 5 heads.
- D
  $(5 8) (0.5)^{8} \approx 0.219$ is wrong; the correct answer is 0.273.
  Why not D: $(5 8) (0.5)^{8} = 56/256 \approx 0.219$ , not 0.273; this is actually the correct formula.
Explanation
$P (X = 5) = (5 8) (0.5)^{5} (0.5)^{3} = 56 \times (0.5)^{8} = 56/256 \approx 0.219$ . $(5 8) = (3 8) = 56$ counts the number of ways to choose which 5 of 8 flips are heads.

Key takeaway
Binomial PMF: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$; the binomial coefficient counts arrangements.
Question 12 · Hard
In a large population, 1% of people have a certain disease. A test has a sensitivity of 95% (true positive rate) and a specificity of 90% (true negative rate). A randomly selected person tests positive. What is the approximate probability they actually have the disease?
- A
  About 95%
  Why not A: 95% is the sensitivity — the probability of a positive test given the disease is present, not the reverse.
- B
  About 87%
  Why not B: Averaging sensitivity and specificity is not how Bayes' theorem works.
- C
  About 8.7%Correct
- D
  About 1%
  Why not D: 1% is the prevalence; the positive test result updates this probability upward, but not to 95%.
Explanation
Use Bayes' theorem. True positives: $0.01 \times 0.95 = 0.0095$ . False positives: $0.99 \times 0.10 = 0.099$ . Total positives: $0.0095 + 0.099 = 0.1085$ . $P (disease ∣ +) = 0.0095/0.1085 \approx 0.0876 \approx 8.7%$ . Low prevalence makes false positives dominate.

Key takeaway
Even with a good test, low disease prevalence causes many false positives; use Bayes' theorem to find the true probability of disease given a positive result.